Question

A single force P acts at Cin a direction perpendicular to the handle BC of the crank shown. Determine the moment My of P about the x-axis when 0 = 65°, knowing that My=-33.000 N·m and Mz=-60.000 Nm. 200 mm 150 mm The moment My of P about the x-axis is N.m

Answer 1

The position vector written as. about c can be r = (loo +150) mm i + 200 mm sinoj + 200 mm coso û = 0.25it 0.2 j sin 6st 0.2 casosk (o.asit 0.18 9 + 0.0848 m acting at c, p = p (-sin & jt cas où) - Fonce Now, m = ² x r t p = 10.25 0.18 0.084 -pbingo pcoso to i :. (Mxit My } t M2R). = 0.18 pcoso + 0.084 p sind i - (0.25 pcos d) j - (o.as psin 6) we can get tand value from My & M₂ as they have sin and cas value.

- - 0.25 p sind му -o-as r casф – tan of i tanda - bo = 20 .: © ten (20) = 61.199 M₂ - - 0.25 p sin & ş.-60 = -0.25 x sin (61.19.) xp ps 0 28 x din 6). 10') 60 = 273.9 N x = = 0.18 p20sº 0.084 pdin ♡ 0.18 x 273.9 x ces (61119) +0.084 273.9 x 8cm 1:19) = - 23.76 + 20.16 43.92 N